Y'' - y' = e^x [2nd order nonhomogenous diff Eq]

[JeffNYC]

I have an equation I need to solve by using undetermined coefficients:

y'' - y' = e^x

The auxiliary equation is:

r²- r = 0 , so 2 real roots (R₁=0, R₂ = 1)

So, y_c(x) = C₁ + C₂e^x

Now for the particular solution:

I can try Ae^x but this is already present in the complementary solution. Do I use:

y_p(x) = xAe^x

Is this the right move at this point in the problem?

 

[rock.freak667]

Yes, since r=1 is a root of the auxiliary equation, your PI should be Axe^x

 

[Mute]

You could also note that you have a first order ODE in disguise: set v = y' and you have

v' - v = e^x

If you multiply by e^(-x) you get

e^(-x)v' - e^(-x)v = (e^(-x)v)' = 1

Which means e^(-x)v = x + C => y' = xe^(x) + Ce^(x); integrate once more to get y.

 

[Marin]

Hi there!

there's also another way of finding the particular solution, but it works, iff the inhomogeneous part is of the form e^(ax) where a is any complex number (i.e. it can be also real or pure imaginary)

so, here's the general formula:
y''+py'+qy=be^{\alpha x}

using the differential operator D=\frac{d}{dx}, we obtain:

(D^2+pD+q)y=be^{\alpha x}

considering D^2+pD+q as the characteristic polynomial and rewriting it as p(D):

p(D)y=be^{\alpha x}

so far so good: Now every particular solution has the following form:

y_p=\frac{be^{\alpha x}}{p(\alpha)}

! IF it turns out that p(\alpha)=0, we take:

y_p=\frac{bxe^{\alpha x}}{p'(\alpha)}

! NOW, if also p'(\alpha)=0, we take:

y_p=\frac{bx^2e^{\alpha x}}{p''(\alpha)}

 

[Marin]

In this example:

y''-y'=e^x
using the differential operator D=\frac{d}{dx}, we obtain:
(D^2-D)y=e^x

Now, consider D^2-D as the characteristic polynomial and write p(D) instead:

p(D)y=e^x

we try the particular solution:

y_p=\frac{be^{\alpha x}}{p(\alpha)}=\frac{e^x}{p(1)}

but 1 is a root of the char. polynomial(i.e. p(1)=0), so we take:

y_p=\frac{bxe^{\alpha x}}{p'(\alpha)}=\frac{xe^x}{p'(1)}

p'(1)=2(1)-1=1 and we obtain for our y_p:

y_p=xe^x

*** Note that trig functions sinx and cosx could be represented as imaginary or real part of the complex exponential - then this formula would also provide us with the correct y_p :)

Marin

 

[JeffNYC]

Great answers all - I didn't realize there were several ways to attack this type of problem.

Jeff